I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?

I understand why f^-1(f(x)) has a domain of x>3, but I don't know why the domain of f(f^-1(x)) is x > 0. Is it because since f(x) has a domain of x >3, so the range of f inverse is also [3, infinity). f inverse = (cube root x) + 3. For the range of f inverse to be [3, infinity), x > 0? Is my reasoning correct?

There is a rule that says:

These rules can be thought of as a consequence of swapping y and x around in y = f(x) to get the rule for the inverse function ..... y's and x's swap --> domain and range swap .....

In your example,

and .

Therefore:

.

.

The composite function g(f(x)) exists iff . When that condition is met, the domain of g(f(x)) is . In you present question, this condition is met for both and .